S*********g
发帖数: 5298 | 1 For minimum nn distance case, we can think the ball has radius x/2.
Then, we have 1-(n-1)x spare space(The two balls on the two ends
can only count one half each). Therefore, the problem becomes
putting n balls of radius 0 into 1-(n-1)x with at least 2 balls on the
same position, this gives the probability
p(x)=n(n-1)(1-(n-1)x)^(n-1)
For the nn distance case, we can think one ball is stick to another
one by a stick of length x. Therefore,the problem becomes putting
(n-1) balls of zero radius int | r*******y
发帖数: 1081 | 2 For the continuous random points, the probability of two points are at
same position is 0, right ?
How to get p(x)=n(n-1)(1-(n-1)x)^(n-1) ?
Thanks a lot.
【在 S*********g 的大作中提到】
: For minimum nn distance case, we can think the ball has radius x/2.
: Then, we have 1-(n-1)x spare space(The two balls on the two ends
: can only count one half each). Therefore, the problem becomes
: putting n balls of radius 0 into 1-(n-1)x with at least 2 balls on the
: same position, this gives the probability
: p(x)=n(n-1)(1-(n-1)x)^(n-1)
: For the nn distance case, we can think one ball is stick to another
: one by a stick of length x. Therefore,the problem becomes putting
: (n-1) balls of zero radius int
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